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| Time Limit: 2 second(s) | Memory Limit: 32 MB |
You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.
For each case, print the case number and N. If no solution is found then print ‘impossible‘.
Sample Input |
Output for Sample Input |
|
3 1 2 5 |
Case 1: 5 Case 2: 10 Case 3: impossible |
题意：给你一个数字，这个数字代表N！后面有几个0。给出这个数字，计算N的值。
解题思路：
任何质因数都可以写成素数相乘的形式。所以计算一个数的阶乘后面几个0，只需计算这个数包含多少5即可。（关于这点不清楚的点：点击打开链接）。
可以用二分法，找出这个点。想到用二分这道题也就没什么难度了。
AC代码;
<span style="font-size:18px;">#include <stdio.h>
#include <math.h>
#include <vector>
#include <queue>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL solve(LL n)
{
LL num=0;
while(n){
num+=n/5;
n/=5;
}
return num;
}
LL er(LL n)
{
LL x=1;
LL y=1844674407370;
LL mid;
LL res=-1;
while(y>=x){
mid=(x+y)/2;
LL ans=solve(mid);
if(ans==n){
res=mid;
y=mid-1;
//return mid;
}
else if(ans>n){
y=mid-1;
}
else if(ans<n){
x=mid+1;
}
}
return res;
}
int main()
{
int t;
scanf("%d",&t);
int xp=1;
while(t--){
LL n;
scanf("%lld",&n);
LL ans=er(n);
if(ans==-1) printf("Case %d: impossible\n",xp++);
else printf("Case %d: %d\n",xp++,ans);
}
return 0;
}
</span>版权声明：本文为博主原创文章，转载请注明出处。
LightOJ Trailing Zeroes (III) 1138【二分搜索+阶乘分解】
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原文地址：http://blog.csdn.net/ydd97/article/details/47816067
